CSOR W4231.002 – Spring, 2019 hw6solns.pdf - CSOR W4231.002 – Spring 2019 hw6 solutions 1 Solution to problem 1 We will reduce this problem to the

# CSOR W4231.002 – Spring, 2019 hw6solns.pdf - CSOR...

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CSOR W4231.002 – Spring, 2019 hw6 solutions 1. Solution to problem 1 We will reduce this problem to the problem of finding a feasible circulation in a flow network with demands. The reduction transformation: given an input instance to our problem, that is, a flow network G = ( V, E ) with edge capacities c e , edge lower bounds e and node demands d ( v ), we construct an input ( G 0 , c 0 , d 0 ) to the circulation with demands problems, where G 0 = ( V 0 , E 0 ) is a flow network with edge capacities c 0 e and node demands d 0 ( v ): (a) set G 0 = ( V, E ); (b) set c 0 e = c e - e , for every e E (c) compute m v = e into v e - e out of v e , for every v V (d) set d 0 ( v ) = d ( v ) - m v , for all v V Clearly the transformation requires polynomial-time. Equivalence of the instances: We will show that there is a feasible circulation f with demands and lower bounds in G if and only if there is a feasible circulation with demands f 0 in G 0 . (= ) Suppose that f is a feasible circulation f with demands and lower bounds in G , that is for every e E , e f ( e ) c e (1) for every v V , f in ( v ) - f out ( v ) = d ( v ) (2) Define f 0 ( e ) = f ( e ) - e . From equation ( ?? ) and (b), 0 f 0 ( e ) c e - e = c 0 e , so f 0 satisfies capacity constraints. Also, f 0 satisfies node demand constraints in G 0 : f 0 in ( v ) - f 0 out ( v ) = X e into v ( f ( e ) - e ) - X e out of v ( f ( e ) - e ) = X e into v f ( e ) - X e out of v f ( e ) - X e into v e - X e out of v e = f in ( v ) - f out ( v ) - m v = d ( v ) - m v = d 0 ( v ) where the last equality follows from equation ( ?? ) and (d). ( =) Conversely suppose that f 0 is a feasible circulation with demands in G 0 , that is for every e E , 0 f 0 ( e ) c 0 e (3) for every v V , f 0 in ( v ) - f 0 out ( v ) = d 0 ( v ) (4) Define f ( e ) = f 0 ( e ) + e . From equation ( ?? ) and (b), e f ( e ) c e , so f satisfies capacity and lower bounds constraints in G . Also, f satisfies node demand constraints in G ; this follows from an entirely similar derivation as above using equation ( ?? ) and (d).

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2. Solution to problem 2 Reduction from SAT : given an arbitrary SAT formula φ with n variables, construct the instance ( φ, n ) of Stingy SAT. The reduction takes polynomial time and φ is a yes instance of SAT if and only if ( φ, n ) is a yes instance of Stingy SAT (that is, φ can be satisfied by setting at most all the variables to 1). 3. Solution to problem 3 Let A be the polynomial time algorithm that on input a formula φ answers yes if φ is satisfiable and no , otherwise. If A ( φ ) = no , then output no and terminate. Otherwise, we know that φ has a satisfying truth assignment. Set x 1 = 1. If A ( φ | x 1 = 1) = no , then set x 1 = 0 (in this case, we know that x 1 = 0 satisfies φ , since φ is satisfiable). Repeat with variables x 2 , . . . , x n , considering them one at a time. After considering all the variables, we have a truth assignment that satisfies φ . This gives a polynomial-time algorithm for finding a satisfying truth assignement, given the hypothetical polynomial-time algorithm A .
4. Solution to problem 4 Decision version: Given an m × n binary matrix A of customers and products, and a target value k , are there at least k orthogonal customers?

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